Answer:
a) 112.5 m
b) 15.81s
Explanation:
a)We can use the following equation of motion to calculate the velocity v of the rocket at s = 500 m at a constant acceleration of a = 2.25 m/s2
[tex]v^2 = 2as[/tex]
[tex]v^2 = 2*2.25*500 = 2250[/tex]
[tex]v = \sqrt{2250} = 47.4 m/s[/tex]
After the engine failure, the rocket is subjected to a constant deceleration of g = -10 m/s2 until it reaches its maximum height where speed is 0. Again if we use the same equation of motion we can calculate the vertical distance h traveled by the rocket after engine failure
[tex]0^2 - v^2 = 2gh[/tex]
[tex]-2250 = 2(-10)h[/tex]
[tex]h = 2250/20 = 112.5 m[/tex]
So the maximum height that the rocket could reach is 112.5 + 500 = 612.5 m
b) Using ground as base 0 reference, we have the following equation of motion in term of time when the rocket loses its engine:
[tex]s + vt + gt^2/2 = 0[/tex]
[tex]500 + 47.4t - 10t^2/2 = 0[/tex]
[tex]5t^2 - 47.4t - 500 = 0[/tex]
[tex]t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
[tex]t= \frac{47.4\pm \sqrt{(-47.4)^2 - 4*(5)*(-500)}}{2*(5)}[/tex]
[tex]t= \frac{47.4\pm110.67}{10}[/tex]
t = 15.81 or t = -6.33
Since t can only be positive we will pick t = 15.81s
